Find the unit tangent vector $T$ and the curvature $\kappa$ for the following parameterized curve.
\[
r(t)=\langle 2 t+1,5 t-7,4 t+14\rangle
\]
\[
T=
\]
(Type exact answers, using radicals as needed.)

Step 1 ：First, we need to find the derivative of the vector function $r(t)$, which is $r'(t)$. This will give us the tangent vector at any point $t$.

Step 2 ：We have $r(t) = \langle 2t+1, 5t-7, 4t+14 \rangle$. So, $r'(t) = \langle 2, 5, 4 \rangle$.

Step 3 ：The unit tangent vector $T$ is obtained by normalizing $r'(t)$. That is, $T = \frac{r'(t)}{||r'(t)||}$, where $||r'(t)||$ is the magnitude of $r'(t)$.

Step 4 ：The magnitude of $r'(t)$ is $||r'(t)|| = \sqrt{(2)^2 + (5)^2 + (4)^2} = \sqrt{4 + 25 + 16} = \sqrt{45} = 3\sqrt{5}$.

Step 5 ：So, the unit tangent vector $T$ is $T = \frac{\langle 2, 5, 4 \rangle}{3\sqrt{5}} = \langle \frac{2}{3\sqrt{5}}, \frac{5}{3\sqrt{5}}, \frac{4}{3\sqrt{5}} \rangle$.

Step 6 ：Next, we need to find the curvature $\kappa$. The curvature is given by $\kappa = \frac{||T'(t)||}{||r'(t)||}$, where $T'(t)$ is the derivative of the unit tangent vector $T$.

Step 7 ：We have $T = \langle \frac{2}{3\sqrt{5}}, \frac{5}{3\sqrt{5}}, \frac{4}{3\sqrt{5}} \rangle$. So, $T'(t) = \langle 0, 0, 0 \rangle$.

Step 8 ：The magnitude of $T'(t)$ is $||T'(t)|| = \sqrt{(0)^2 + (0)^2 + (0)^2} = 0$.

Step 9 ：So, the curvature $\kappa$ is $\kappa = \frac{||T'(t)||}{||r'(t)||} = \frac{0}{3\sqrt{5}} = 0$.

Step 10 ：Therefore, the unit tangent vector $T$ is $\boxed{\langle \frac{2}{3\sqrt{5}}, \frac{5}{3\sqrt{5}}, \frac{4}{3\sqrt{5}} \rangle}$ and the curvature $\kappa$ is $\boxed{0}$.